Question

Describe the preparation of 1 L of 0.100 M acetate buffer, pH 5.3, if you have 0.100 M stock solution of acetic acid, pKa 4.52. You also have stock solution of 1.00 M NaOH and 1.00 M HCl. Determine whether you need NaOH or HCl to make the buffer and how much should be added?

Answer 1

pH = pKa + log {[CH3COO-]/[CH3COOH]}
5.3 = 4.52 + log {[CH3COO-]/[CH3COOH]}
log {[CH3COO-]/[CH3COOH]} = 0.78
[CH3COO-]/[CH3COOH] = 6.03
[CH3COO-] = [CH3COOH] * 6.03

Since you need CH3COO- to be formed, you need to add NaOH

number of moles of CH3COO- / number of moles of CH3COOH = 6.03
n1/n2= 6.03
n1 = 6.03*n2

also total number of moles = moarity of acetate buffer * volume
     = 0.1 M * 1L
     = 0.1 mol

n1+n2 = 0.1 mol
6.03*n2 + n2 = 0.1
n2 = 0.014 mol

volume of CH3COOH = n2/M of CH3COOH = 0.014 / 0.1 = 0.14 L

n1 = 6.03*n2 = 6.03*0.014 mol = 0.086 mol
volume of CH3COO- = n1/M of CH3COO- = 0.086 / 1 = 0.086 L

Remaing water should be added

volume of water = 1 - 0.14 - 0.086 =0.774 L